Find the sum of the A.P. 7, 101/2, 14, …, 84.

MathsCBSE Class 10 · Basic · Board exam 2025, set 430/4/1 · 3 marks · Short answer

CBSE Class 10 Mathematics Basic 2025: question 27(a)

Set in the CBSE Class 10 board exam of 2025, paper 430/4/1, section C, question 27(a), for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q27(a) · 3 marks · Short answerArithmetic Progressions

Find the sum of the A.P. $7, 10 \frac{1}{2}, 14, \dots, 84$ .

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 430/4/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Write $10 \frac{1}{2}$ as $\frac{21}{2}$ and find the common difference.

Hint 2

Use $a_{n} = a + (n - 1) d$ with $a_{n} = 84$ to find how many terms there are.

Hint 3

Then use $S_{n} = \frac{n}{2} (a + l)$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

$a = 7$ , $d = \frac{21}{2} - 7 = \frac{7}{2}$ , last term $l = 84$
1 mark
$a_{n} = a + (n - 1) d$ : $84 = 7 + (n - 1) \times \frac{7}{2} \Rightarrow (n - 1) \times \frac{7}{2} = 77 \Rightarrow n - 1 = 22 \Rightarrow n = 23$
1 mark
$S_{23} = \frac{n}{2} (a + l) = \frac{23}{2} (7 + 84) = \frac{23 \times 91}{2} = \frac{2093}{2}$ , i.e. $1046.5$
1 mark

Answer: Sum $= \frac{2093}{2} = 1046.5$

Where the marks are

Identifies $a = 7$ , $d = \frac{7}{2}$ (1 mark)
Finds $n = 23$ (1 mark)
Finds $S_{23} = \frac{2093}{2}$ (1 mark)

Revise the chapter

Chapter: Arithmetic Progressions (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Arithmetic Progressions, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 430/4/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2025: question 27(a)

Hints

Step-by-step solution

Revise the chapter

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